Question

A ball is thrown vertically upward a returns to the same height in 5.37 s.   A second ball is thrown at 37 degrees . What speed must the second ball have to reach the same height as the first ball?

Answer 1

Time = 5.37s
Time to reach max. = 5.37/2 = 2.68 sec .
So it takes 2.68 sec . to reach its max. height. Using that we can now calculate how far downwards the ball fell FROM the max. This is much easier than trying to calculate the max. height upwards. So at the max. point:

u=0
a=9.8 (+ve since we can take the downwards direction to be +ve)
t=2.68 sec .
s=unknown

Now, using s=ut+0.5*a*t^2 (s=displacement, u=initial v, a=acceln. and t=time)

s=0+0.5*(9.8)*2.68 sec ^2
s= 35.2 m

So, the ball fell 35.2 m, using symmetry we can therefore assume that its max. was also 35.2 m

Now we can forget apart ball 1 and move onto ball 2. What's important to understand here the independence of the horizontal and vertical components. If its been thrown at 56 degrees to the horizontal then, then we have:

u= usin37 (This is the vertical component)
u= ucos37 (This is the horizontal component)

Now, we know that we want our maximum to be 35.2 m, we know acceln. vertically is -9.8 (-ve since upwards is positive and g acts downwards). So we have information necessary to use:

v^2=u^2+2as

So substituting in our known information:

0=(usin37)^2+2*-9.8*35.2 m(v=0 since at its maximum point, the balls v=0)
0=(usin37)^2-691
691-=usin37^2
sqrt(691)=usin37
26.2/sin37=u
Therefore u= 71 m/s